 Hi and welcome to the session. Let us discuss the following question. Question says evaluate the following definite integrals as limit of sums. Given integral is from 2 to 3 x square dx. Clearly we can see 2 is the lower limit of the integral and 3 is upper limit of the integral. First of all let us understand that according to the definition of definite integral as limit of the sum definite integral from a to b fx dx is equal to b minus a multiplied by limit of n tending to infinity 1 upon n multiplied by fA plus fA plus h plus fA plus 2h till fA plus n minus 1 multiplied by h. Now in this expression h is equal to b minus a upon n. Now this expression is our key idea to solve the given question. Let us now start with the solution. Now we have to evaluate integral from 2 to 3 x square dx as limit of sum. Now comparing this integral with integral given in key idea we get fx is equal to x square, a is equal to 2, b is equal to 3. So here we can write a is equal to 2, b is equal to 3, fx is equal to x square. Now let us find out h is equal to b minus a upon n which is further equal to 3 minus 2 upon n or we can say it is equal to 1 upon n. So we get h is equal to 1 upon n. Now using key idea we can write definite integral from 2 to 3 x square dx is equal to b minus a is equal to 1. So here we will write 1 multiplied by limit of n tending to infinity 1 upon n multiplied by fA and a is equal to 2. So here we will write f2 plus f2 plus h plus f2 plus 2h plus till f2 plus n minus 1 multiplied by h. Now we know h is equal to 1 upon n. So this expression is equal to limit of n tending to infinity 1 upon n multiplied by square of 2 plus square of 2 plus 1 upon n plus square of 2 plus 2 upon n plus square of 2 plus n minus 1 multiplied by 1 upon n whole square we know we are given fx is equal to x square. So f2 is equal to 2 square f2 plus h is equal to 2 plus 1 upon n whole square we know value of h is 1 upon n. Similarly here f2 plus 2h is equal to 2 plus 2 upon n whole square and f2 plus n minus 1 multiplied by h is equal to 2 plus n minus 1 multiplied by 1 upon n whole square. Now we will simplify this bracket we know square of 2 is 4. So here we will write 4. Here we will apply the formula of a plus b whole square. So we will write a square plus b square plus 2ab that is 4 upon n. Now here also we will apply formula of a plus b whole square. It is equal to a square plus b square plus 2ab that is 8 upon n. Now here again we will apply formula of a plus b whole square it is equal to a square plus b square. Here b square is n minus 1 upon n whole square plus 2ab that is 4 multiplied by n minus 1 upon n. Clearly we can see this 2 square is occurring n minus 1 number of times and this 4 is also 2 square. So total number of 2 square occurring in this bracket is n. So we can write this expression is equal to limit of n tending to infinity 1 upon n multiplied by 4n plus square of 1 upon n plus square of 2 upon n plus square of n minus 1 upon n plus 4 upon n plus 8 upon n plus 4 multiplied by n minus 1 upon n. In this step we had written these terms in one bracket and these terms in one bracket. Clearly we can see 1 upon n square is common in all these terms. So we will take 1 upon n square common in this bracket. We get limit of n tending to infinity 1 upon n multiplied by 4n plus 1 upon n square multiplied by 1 square plus 2 square plus n minus 1 whole square. And here in these terms we can see 4 upon n is common. So we will take 4 upon n common in this bracket and we get 4 upon n multiplied by 1 plus 2 plus n minus 1. Now this expression is further equal to limit of n tending to infinity 1 upon n 4n plus 1 upon n square multiplied by n multiplied by n minus 1 multiplied by 2n minus 1 upon 6. This is a GP and sum of this geometric progression is equal to n multiplied by n minus 1 multiplied by 2n minus 1 upon 6. And this series is an AP. So sum of this series is equal to n multiplied by n minus 1 upon 2. Now simplifying this expression further we get limit of n tending to infinity 1 upon n 4n plus n minus 1 multiplied by 2n minus 1 upon 6n. Here this n will get cancelled by 1n in the denominator and we are left with n minus 1 multiplied by 2n minus 1 upon 6n. Now in this term we will cancel common factor 2 from numerator and denominator both and this n will get cancelled by this n. And we are left with 2 multiplied by n minus 1. Now this expression is further equal to limit of n tending to infinity 1 upon n multiplied by 4n plus n minus 1 multiplied by 2n minus 1 is equal to 2n square minus n minus 2n plus 1 upon 6n and multiplying these two terms we get 2n minus 2. Now adding these two terms we get 6n so here we can write limit of n tending to infinity 1 upon n multiplied by 6n plus 2n square minus 3n plus 1 upon 6n minus 2. Now multiplying each term by 1 upon n we get limit of n tending to infinity 6 plus 2n square minus 3n plus 1 upon 6n square minus 2 upon n. Now this is further equal to limit of n tending to infinity 6 plus 2n square upon 6n square minus 3n upon 6n square plus 1 upon 6n square minus 2 upon n. n square will get cancelled by n square here 1n will get cancelled by this n present in the numerator. Now this limit is equal to 6 plus 2 upon 6 minus 0 plus 0 minus 0. Now we can cancel common factor 2 from numerator and denominator both here and we get 6 plus 1 upon 3. Now adding these two terms by taking their LCM we get 19 upon 3. So we get given integral from 2 to 3 x square dx is equal to 19 upon 3. This is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.